Home
Class 11
PHYSICS
A ball is thrown upwards from the top of...

A ball is thrown upwards from the top of a tower `40 m` high with a velocity of `10 m//s.` Find the time when it strikes the ground. Take `g=10 m//s^2`.

Text Solution

Verified by Experts

In the problem `u= + 10 m//s, a =-10 m//s^(2) and s =- 40 m ` ( at the point where ball strikes the ground)
Substituting in` s = ut +(1)/(2) at^(2)`
`-40= 10t - 5t^(2) or 5t^(2) -10t -40=0 or t^(2) - 2t -8 =0`
Solving this we have t= 4 s and -2 s. Taking the positive value t = 4s.
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN|Exercise BEGINNER S BOX-1|6 Videos

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN|Exercise BEGINNER S BOX-2|7 Videos

  • KINEMATICS

    ALLEN|Exercise Integer Type Question|3 Videos

  • MISCELLANEOUS

    ALLEN|Exercise SUBJECTIVE QUESTION|9 Videos

Similar Questions

Explore conceptually related problems

A stone thrown upwards from the top of a tower 85 m high, reaches the ground in 5s. Find the greatest height above the ground.Take g = 10 ms^(-2)

A ball is thrown vertically upwards from the top of tower of height h with velocity v . The ball strikes the ground after time.

A ball is thrown vertically upwards from the top of a tower of height 60 m with a speed of 20 m//s. (i) The ball strikes the ground after 6 s (ii) The ball strikes the ground with speed 40 m//s (iii) The distance of ball above the ground after 5 s is 35 m The maximum height attained by ball above the ground is 80 m

A ball is thrown from the top of a tower with an intial velocity of 10 m//s at an angle 37^(@) above the horizontal, hits the ground at a distance 16 m from the base of tower. Calculate height of tower. [g=10 m//s^(2)]

A ball is thrown vertically upwards from the top of a tower with a velocity of 10m/s. If the ball falls on the ground after 5 seconds, the height of the tower will be? (use g=10m/ s^(2) )

A ball is thrown from the top of a tower with an initial velocity of 10 m//s at an angle of 30^(@) above the horizontal. It hits the ground at a distance of 17.3 m from the base of the tower. The height of the tower (g=10m//s^(2)) will be

A ball is thrown upward with speed 10 m/s from the top to the tower reaches the ground with a speed 20 m/s. The height of the tower is [Take g=10m//s^(2) ]

A ball is thrown horizontally with a speed of 20 m//s from the top of a tower of height 100m velocity with which the ball strikes the ground is .

An object is thrwon vertically upwards with a velocity of 60 m/s. After what time it strike the ground ? Use g=10m//s^(2) .

A particle is projected upwards from the roof of a tower 60 m high with velocity 20 m//s. Find (a) the average speed and (b) average velocity of the particle upto an instant when it strikes the ground. Take g= 10 m//s^2